Elizabeth lives in San Francisco and works in Mountain View. In the morning, she has $3$ transportation options (take a bus, a cab, or a train) to work, and in the evening she has the same $3$ choices for her trip home. If Elizabeth randomly chooses her ride in the morning and in the evening, what is the probability that she'll use a cab exactly one time?
Solution: $\text{Probability} = \dfrac{\text{Favorable outcomes}}{\text{Total possible outcomes}}$ There are $3$ possible choices for each trip, so there are $3\times3=9$ total ways she can get to work and back. If Elizabeth chooses randomly, each outcome is equally likely. The green rows show the outcomes where Elizabeth uses a cab once. There are $4$ favorable outcomes. Morning Evening Bus Bus ${\text{Bus}}$ ${\text{Cab}}$ Bus Train ${\text{Cab}}$ ${\text{Bus}}$ Cab Cab ${\text{Cab}}$ ${\text{Train}}$ Train Bus ${\text{Train}}$ ${\text{Cab}}$ Train Train The probability that Elizabeth will use a cab once is $4$ out of $9$, or $\dfrac49$.